# Characterization of Boundary by Basis

## Theorem

Let $T = \struct {S, \tau}$ be a topological space.

Let $\BB \subseteq \tau$ be a basis.

Let $A$ be a subset of $T$.

Let $x$ be a point of $T$.

Then $x \in \partial A$ if and only if:

for every $U \in \BB$:
if $x \in U$
then $A \cap U \ne \O$ and $\relcomp S A \cap U \ne \O$

where:

$\relcomp S A = S \setminus A$ denotes the complement of $A$ in $S$
$\partial A$ denotes the boundary of $A$ in $T$.

## Proof

### Sufficient Condition

Let $x \in \partial A$.

Let $U \in \BB$.

By definition of basis, $U$ is an open set of $T$.

if $x \in U$
then $A \cap U \ne \O$ and $\relcomp S A \cap U \ne \O$.

$\Box$

### Necessary Condition

Let $x$ be such that for every $U \in \BB$:

if $x \in U$
then $A \cap U \ne \O$ and $\relcomp S A \cap U \ne \O$.

By Characterization of Boundary by Open Sets, to prove that $x \in \partial A$ it is enough to prove that:

for every open set $U$ of $T$:
if $x \in U$ then $A \cap U \ne \O$ and $\relcomp S A \cap U \ne \O$.

Let $U$ be an open set of $T$.

Let $x \in U$.

By definition of (analytic) basis, there exists $V \in \BB$ such that:

$x \in V \subseteq U$

By assumption:

$A \cap V \ne \O$

and:

$\relcomp S A \cap V \ne \O$
$A \cap V \subseteq A \cap U$

and:

$\relcomp S A \cap V \subseteq \relcomp S A \cap U$

So:

$A \cap U \ne \O$ and $\relcomp S A \cap U \ne \O$

and hence the result.

$\blacksquare$